In my math class, we are working on Geometric Optimization problems. We have to create an equation, and then solve for one variable, in terms of another variable. Then, using an expression, we find the vertex (by converting it to an equation) and then use the x-value of the vertex to get both variables in the result. For example, in this problem:

A rectangular field is to be fenced off next to a river.
The budget for building the fence is \$1,800.
No fence is required along the river.
The material to be used for the side of the field parallel to the river costs \$5 per foot.
The material to be used for other two sides costs \$3 per foot.
Find the dimensions of the field with the greatest possible area.

You would solve it by creating an equation, $1800=5x+6y$, and then solve for $y$ in terms of $x$: $y=\frac{1800-5x}6$. Then, we would make an expression, of the area: $xy$ which is $\frac{x(1800-5x)}6$. We would plug this into a graphing calculator (or do it by hand), and the $x$-value of that result is one of the dimensions. We can find the other by substituting in the $x$-value of the vertex into $y=\frac{1800-5x}6)$, where $y$ is the other dimension. The answer would be 150 by 180, because this produces the greatest area.

Now, on to the problem I am lost on:

An ordinary soda can has volume of $355\ \mathrm{cm}^3$. Assume that soda cans are cylinders. A typical goal for designing a can would be to minimize the surface area (in order to minimize the amount of material needed). Find the radius and height of a soda can with the least possible surface area.

So, I started off with doing $355=\pi r^2h$ (formula for volume of a cylinder), and solved for $h$: $h=\frac{355}{\pi r^2}$. Then, I took the formula for the surface area of a cylinder: $2\pi rh+2\pi r^2)$, and then substituted in the value of $h$: $$2\pi r\left(\frac{355}{\pi r^2}\right) + 2\pi r^2$$

I plugged this expression into my calculator (replacing $r$ with $X$), to find this:

This has no minimum, so can anyone please help me?

I looked online, and the answers are radius: $3.84\ \mathrm{cm}$, height: $7.68\ \mathrm{cm}$, but I have no idea how my teacher got them, since no explanation was provided.

I'm only a freshman in high school, so please don't suggest things that are somewhat complex.

  • 2
    $\begingroup$ Only the graph for $r > 0$ is relevant, and that has a minimum, one can even recognize it on your graph. – Are you familiar with derivatives, and how to use them to solve extremal problems? $\endgroup$ – Martin R Jan 9 '16 at 16:14
  • $\begingroup$ @MartinR $r>0$ was a part of the issue, I just also needed to include the dip in the "guess" on the calculator, which I didn't do. $\endgroup$ – Flare Cat Jan 9 '16 at 16:23

Nevermind, please ignore my question, I figured it out. I had to make sure the X was greater than 0, and I included the dip in the calculator in the guess.

Edit: After answering my own question and reloading the page, I realized @MartinR said r had to be greater than 0, which was the issue. So, good job on figuring it out, Martin.


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